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February 14, 2014, 02:44 |
Turbulence
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#1 |
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Geth
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Why free shear flows become turbulent at very low Re?..
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February 14, 2014, 05:02 |
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#2 |
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Filippo Maria Denaro
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February 14, 2014, 05:29 |
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#3 |
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Geth
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am asking..like a smoke from cigeratte is becomes turbulent at very low Re..but for flat plate it is very high...
Also, trans. Re is in the range of 10 power 5 for flat plate but in terms of 1000 for flow inside pipe..why?..can you give me clear description about these three things |
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February 14, 2014, 05:38 |
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#4 | |
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Filippo Maria Denaro
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Quote:
But transition to turbulence is due to several factors...in pipe you have transition at Re = O(10^3), on an airfol at Re=O(10^5) you have still laminar flow. Turbulence can exists at low Re number, for example flow from a small orifice becomes rapidly turbulent. Smoke from cigarette is just a visualization of turbulence already existing in air, is not the case of jet stream or mixing layer... |
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February 14, 2014, 08:12 |
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#5 |
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Geth
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Ok..what is reason for these two above examples..as Re less for pipe and as same Re we get laminar in airfoil..
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February 14, 2014, 09:04 |
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#6 |
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Alex
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Reynolds numbers for different types of flows can not be compared like that.
If the Reynolds number for the flow over an airfoil is 10 times higher than the Reynolds number number for the flow in a pipe, this does not imply that the airfoil flow has to be 10 times "more turbulent" than the pipe flow. The Reynolds number can only be used to compare similar flows, like the flow through the same pipe at two different velocities. The value of the critical Reynolds number at which a flow becomes turbulent is quite arbitrary and can not be compared to a different kind of flow. As a last example, we could simply define a new Reynolds number for the flow in a circular pipe and say it is where u is the average velocity, r is the pipe radius and nu is the kinematic viscosity. Since our definition yields a Reynolds number lower by a factor of 2 compared to the usual definition of a Reynolds number for pipe flows, its critical value is shifted to ~4600 although the flow is exactly the same. The point is that the definition of the Reynolds number is to some extent arbitrary for each individual flow problem and is not an absolute measure for turbulence. |
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February 14, 2014, 09:23 |
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#7 |
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Geth
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Thank you..
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February 14, 2014, 15:06 |
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#8 |
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Lefteris
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I agree with everything else you said but I don't sure about this one. You can estimate the critical Reynolds number using the Orr-Sommerfeld equation to constract the stability curves.
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Lefteris |
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February 14, 2014, 15:50 |
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#9 |
Senior Member
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It is actually the Reynolds experiment you are talking about. At some distance from a cigarette the Reynolds number (based on the distance from the cigarette, indeed) equals the critical value and the flow becomes turbulent.
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February 15, 2014, 21:19 |
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#10 | |
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Lefteris
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Quote:
Factors that affect transition to turbulence are: a. freestream pressure gradient b. freestream turbulence c. surface roughness d. surface curvature e. surface temperature f. compressibility
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Lefteris |
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February 16, 2014, 11:12 |
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#11 | |
Senior Member
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Quote:
If you look to these two problems more carefully, you will see that transition reynolds number are actually of the same order for both.... Only for the flat plate flow one should consider reynolds number based on the boundary layer thikness. |
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February 17, 2014, 02:40 |
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#12 |
Super Moderator
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Critical reynolds number depends on many factors such as turbulence intensity, roughness etc. so it is random number and its value may deviate by order of magnitude.
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February 17, 2014, 15:50 |
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#13 | ||
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Lefteris
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Quote:
Quote:
I just don't really understand the word "random" since there are tools (equations) for predicting the critical Reynolds number.
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Lefteris |
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February 17, 2014, 16:00 |
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#14 |
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Filippo Maria Denaro
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I think that the initial question was different... free shear turbulent flows at low Re number.
It is a classical example to think about a flow going through a grid and developing a homogeneous isotropi turbulence at some distance. That turbulence exists also at very small Re_lambda (Taylor microscale). |
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February 17, 2014, 16:56 |
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#15 |
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Hamid Zoka
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Dear All;
Let me come back to the first question: "Why free shear flows become turbulent at very low Re?.." I try to give a response by explaining two flow cases: First, consider the flow near a solid wall. there is always a shear stress at near wall region that eventually induces turbulence to the flow. but the wall itself is rigid. so it can not excite the flow in wall normal direction. Therefore the wall impact is limited to the so called shear stress and normal components of Reynolds stress tensor diminish at near wall regions. Second, consider that the instead of wall there is another fluid zone. what happens then? a shear stress develops at the fluid-fluid interface. since the interface is not rigid, so it can be easily deformed by the so called shear stress and as a result some fluid particles in macroscopic scales can go through the fluid-fluid interface and then come back to their zones again. this instability will be soon intensified to form a vortex flow between the fluid zones. therefore this intense mixing can induce the turbulence at vicinity of the interface very quick. note that unlike the first case there is no wall to damp normal components of Reynolds stress tensor and thus turbulence is freely developed at even low Re numbers. Regards |
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